Let $E$ be an elliptic curve over a non-algebraically closed field $K$ and $\overline{K}$ be its algebraic closure. Then for any divisor of degree $1$ on $E(K)$, there exists exactly one $\overline{K}$-rational point $P$ belonging to the linear system, and this point must be a $K$-rational point: Because any subscheme of $E$ contains at least one point of $E$ set-theorically.
To define the degree of a divisor, there might be not enough $K$-rational points, but you can do this by taking an algebraic closure and count the number of $\overline{K}$-rational-points contained in the linear system $|D|$